I am asked to prove that if $ K \leq L$ is a finite field extension of degree $n$ , $f\left(t\right) \in K\left[t\right]$ is irreducible of degree $d > 1$ and $gcd(n,d) = 1$ then $f\left(t\right)$ has no roots in $L$.

My approach was to suppose that there is a root $\alpha$ of $f\left(t\right)$ in $L$, the consider the field extensions:

$$K \leq K(\alpha) \leq L$$

Here I'm pretty sure the order of the extensions is correct as $$K, \{\alpha\} \subset L \Rightarrow K(\alpha) \subset L$$

Then $|K(\alpha):K| = $ deg$f_{\alpha}\left(t\right)$ where $f_{\alpha}\left(t\right)$ is the minimal polynomial of $\alpha$ over $K$

Noting that $f_{\alpha}\left(t\right) \mid f\left(t\right)$ and remembering irreducibility of $f\left(t\right)$ over $K$ we conclude that $af_{\alpha} \left(t\right) = f\left(t\right)$ for some $a \in K$

Thus, $|K(\alpha) : K| = d$

By the Tower Law: $$ |L:K| = |L:K(\alpha)| |K(\alpha):K| $$

$\Rightarrow d \mid n$, a contradiction.

From this I concluded that there cannot exist such a root $\alpha$ of $f\left(t\right)$ in $L$.

I wanted to ask if my proof is correct, because it almost feels too simple to be true and I feel as if I have missed something important. Thank you for any help you might be able to give!

  • 1
    $\begingroup$ It looks good for me. I would probably put $a f_{\alpha} \left(t\right) = f\left(t\right)$ where $a \in K$, since $f(t)$ must not be monic. But this does not change the rest of the proof. $\endgroup$
    – A. Smith
    Oct 15, 2017 at 21:03
  • $\begingroup$ @A.Smith, yes you are right! Thank you! $\endgroup$
    – user366818
    Oct 15, 2017 at 21:04


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